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Full text of "Treatise On Analysis Vol-Ii"

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The inequalities ( and ( express that the linear mappings
y\-+xy and x^>xy are continuous on A with respect to its topology as a
prehilbert space. But it should be carefully noted that in general A is not a
normable algebra for this topology (15.1.8): that is to say, in general the map-
ping (X y)\-+xy is not continuous.

Finally, we have
(    The relation x*x = Qfor xeA implies that x = 0.

For it follows from ( that (xy\xy) = (x*xy\y) = Q, hence
xy = 0 for all y e A. Hence, by (, (x \ yz) = (xz* | y) = 0 for all y and
all z, and by ( this implies that x = 0.

In particular, the right (resp. left) annihilator of A consists only of 0, and
for each x ^ 0 in A the left ideal A.X is nonzero.

In the next two sections we shall determine the structure of two of the most
important types of Hilbert algebras.


(a)    Show that on the algebra A = Mn(C) of all n x n matrices with complex entries,
the only linear forms / such thaif(XY) =f(YX) for all pairs of matrices X, Y are
scalar multiples X\-*plr(X) of the trace.

(b)    Deduce that, if H is an infinite-dimensional separable Hilbert space, there exists
no trace on the algebra ^(H) or on the algebra ^2(H). (Consider a compact operator
U such that U(en) = An <?, where the en form an orthonormal basis of H and the An are
real numbers > 0 such that ^ XI converges but ^ AM does not. Use (a) by considering

n                                                n

the restriction of a hypothetical trace/to the algebra of endomorphisms of the vector
subspace of H generated by the ek with k <i n.)


Throughout this section we shall assume that the Hilbert algebra A is
complete (and thus a Hilbert space) with respect to the norm ||x|| = (x\x)1/2.
We shall also assume that the bilinear mapping (x, y)^-^xy of A x A into A
is continuous with respect to this norm (it can be shown (Section 12.16,
Problem 8c)) that this is in fact a consequence of the other hypotheses). Every
closed self-adjoint subalgebra B of A is a complete Hilbert algebra.